JEE Main 2020PhysicsAlternating CurrentMediumMCQ

JEE Main 2020 — Alternating Current Question with Solution

JEE Main 2020 (09 Jan Shift 2)

Question

In LC circuit the inductance L=40 mH and capacitance C=100 μF. If a voltage Vt=10sin314t is applied to the circuit, the current in the circuit is given as:

Choose an option

Show full solutionCorrect option: A
Correct answer
A0.52cos⁡314t

Step-by-step explanation


z=R2+XC-XL2
R=0
Z=XC-XL
=1ωC-ωL
=1314×100×10-6-314×40×10-3
=31.84-12.56
=19.28 Ω

i=V0Zsin⁡314t+π2
∴i=V0Zcos⁡314t⇒i=1019.28cos⁡314t
⇒i=0.52cos⁡314t

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About this question

This is a previous-year question from JEE Main 2020, covering the Alternating Current chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.